Free hanging chain problem

Once upon a time, when I was still a student, sitting at one boring lecture, I thought about how often a freely hanging rope or chain of a given length can oscillate in one plane and what its shape will be if the oscillations are small. I remember that I solved this problem, but now, after many years, I have already forgotten the details of how I did it. However, it became interesting for me to restore this solution in as much detail as possible and share it with everyone who would be interested. What came of this, read under the cut.



Suppose we have a chain of length l and mass M, suspended by one end, as shown in the figure. Here we will assume that the chain is homogeneous and frictional forces can be neglected. Let's construct a coordinate system in such a way that the origin of coordinates coincides with the suspension point, the X-axis is directed downward, and the Y-axis, perpendicular to the X-axis, will be responsible for the chain deviation from the vertical. In fact, it is necessary to define the function Y (x, t).







To find Y (x, t), let's write down the forces acting on a small section of the chain as shown in the following figure.







The figure shows that the tensile force T is tangent to the chain. Therefore, the tangent of the tilt angle T to the X-axis will be equal to the derivative dY (X) / dX. It is known that if the fluctuations are small, then the tangent is approximately equal to the angle itself in radians. The tensile force T can be calculated using the formula







where l is the length of the chain, g is the acceleration due to gravity, and the







mass per unit length of the chain.



Let us write the equation proceeding from Newton's second law







On the right side of the equation, substitute the value of the tension T while without the corresponding coefficient







Replace the value of the derivative at the point x + dx through the second derivative







Expand the brackets







and cancel the corresponding terms, removing also the term of the second order of smallness.







Substitute the resulting formula into the equation of motion.







Reduce by dx and specific gravity.







Note that this equation does not depend on the specific gravity, therefore, all ropes and chains of equal length will vibrate in the same way, regardless of the mass. In order to solve this equation, we will look for a solution in the form







Substituting it into the equation of motion, we obtain







Dividing it by g and the function itself, we obtain that one part depends only on time, and the other only on X. Therefore, they can be equated some constant.







Let us first consider the part that depends only on X







To solve this equation we will make a change of variable







Then the first derivative will take the following form







and the second derivative will take the following form and the







equation will be rewritten in the form







It is easy to see that this equation can be rewritten in the form







Since it is not yet clear what this equation is, we will try to reduce it to some well-known differential equation.

To do this, we will make the change







In this case, the first derivative will take the following form







and the equation itself is as







follows Move n squared from under the derivative







and cancel it







Make differentiation and get the following equation







We choose n in such a way that there is no free variable at the highest derivative.







We







get the following equation Multiply by 4 and z squared and we get







This is already similar to the well-known Bessel equation, you just need to get rid of the factor from the function itself. To do this, we make another transformation of the variable







In this case, the first derivative will become equal







and the second derivative







Substituting into the equation, we get







If we take







then we get the zero-order Bessel equation







The solution of such an equation has the form







where A and B are constants, and J and Y are zero-order Bessel functions. Substituting the variable z back, we get







After substituting the variable u, we have the following solution







and, finally, returning to the variable x, we







use the fact that our function must be finite at the point x = l. Since the function Y (x) is infinite at zero, B must equal zero and our solution will have the following form







Now we will use the condition that at the suspension point the value of our function must equal zero, that is, y (0) = 0.

It follows from this that







where j are the zeros of the zero-order Bessel function. From here, you can determine the value of the







lambda.Substituting the labda, we get







What after reduction gives its own functions.







Let us give graphs for the first five.















Let us now return to that part of the initial equation, which is responsible for the dependence on time. Knowing the lambda values, you can calculate the natural frequencies







By extracting the root, we get the







Corresponding periods will be equal







Compare this expression with the period of oscillations of a mathematical pendulum.



This concludes our study of the oscillations of a freely hanging chain. Thank you for attention.